more from the 1970s, another one about clocks

quote:

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Regular as Clockwork

Our old watchmaker works weekdays 9.30am to 5pm as regular as ,well, clockwork. I recently took there to be regulated two '8-day' striking clocks - the sort which fully-wound will go nearly 8 days before stopping; they were keeping different times and each was wrong by an exact number of minutes per day, i.e. less than an hour in either case.

He immediately wound the clocks fully, set them to the right time (which was an exact number of minutes after the hour) and put them up on a shelf for observation.

The next Monday, when he went to take down the clocks to start regulating them, he found both of them just starting to strike 8 o'clock simultaneously, which was some hours plus an exact number of minutes past the correct time.

What day and exact time was it when he originally set them?

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gtho4,

If only your old watchmaker kept his shop open untill 6.00pm then this would not have been so difficult???

I reckon the following to your two clocks????

One of your clocks ( Clock A ) is running 2 minutes fast in every hour.

Your other clock ( Clock B ) is running 8 minutes slow in every hour.

Whether by coincidence or not, your friendly watchmaker went to check the clocks exactly 6 days after he had wound and set them to the correct time.

He did this the previous Tuesday at 15:12 hours.???????

Kindest Regards

Bannockburn

mate,
I got a nose bleed when I had a go at this, and surrendered, rather than bleed to death - that's not the right answer - continue reading the italics below if you want a few hints or words of confirmation about the answer, else stop here if you'd like to have another go
cheers,
gt

[ btw anybody is free to have a go at this ]

~~~~~~~~~

the opening hours are indeed relevant, as you've observed, and so is the period of time (it must be less than 8 days), which you've also picked up on - using clock A as fast by X minutes and clock B as slow by Y minutes then, by playing with the maths, both X and Y must be multiples of 5 - the answer is the am of Monday rather than pm of Tuesday

gtho4,

"Mates" we are, and am sure we know each other well enough that this message will be taken solely as an observation and not in way a criticism of either yourself or The Sunday Times.

I am as fallible as anybody but have checked and double-checked my solution and in the last 24 hours have asked a friend or two to check my arithmetic and so far everyone agrees that it is a correct solution. I say this in the context that when I eventually settled down to "bleed" like yourself I continually thought, although I couldn't prove it, that this problem was likely to have more than one solution and am going to put my kneck on the block and suggest that this is the case.

I am not suggesting for one minute that your and or The Sunday Time's solution is wrong but would welcome the opportunity to check it, if only to prove, assuming I am correct, that there are at least two solutions.

Might I beg your indulgence by endeavouring to prove my solution. It is a proof in reverse.

GENERAL:

Let us assume that the watchmaker does in fact wind and set the clocks to the correct time at 15:12 on the Tuesday and returns to check them exactly 6 days ( 144 hours ) later, ie at 15:12 the following Monday. This fully satisfies the question restraints that the shop is only open weekdays from 9:30 to 5:00 and that they are both 8 day clocks.

If it now can be shown that both a clock ( A ) running 2 minutes fast and a clock ( B ) running 8 minutes slow will both be striking 8 o'clock at 15:12 on the following Monday then the case has been proven as the question does not stipulate whether or not on the watchmakers inspection this is the first occasion that this situation has arisen. Indeed it might well be possible although I haven't proved it that they did coincide in such a manner earlier but then again that could have been outwith shop hours. Indeed there is no restraint within the question as to when the watchmaker carried out his inspection other than that it was during shop hours on the following Monday.

PROOF:

At 15:12 on the following Monday the clocks will read as follows:

Clock "A":

15:12 + ( 2 X 144 )mins = 15:12 + (4hrs 48mins) = 20:00

At 15:12 Clock "A" is reading 8 o'clock that evening.

Clock "B":

15:12 - ( 8 X 144 ) mins = 15:12 - (19hrs 12mins) = 20:00

At 15:12 Clock "B" is reading 8 o'clock the previous evening

The arithmetic must be correct, because in any hour with the clocks going 2 minutes fast and 8 minutes slow respectively then they are diverging from each other by 10 minutes in any hour.

Over a period of 6 days ( 144 hours ) the clocks will therefore have diverged by 1440 minutes which is in fact 24 hours.

Please don't tell me there is a mistake in this??

Going straight off for a blood transfusion.

Kindest Regards
Bannockburn

gtho4,

I am an idiot.

This will teach me to read a question more carefully.

The clocks are to be out by a certain number of minutes in a day and not an hour and by no more than an hour in a day.

I hide my head in shame.

Might have another look at it if time permits.

Sincerest apologies

Bannockburn

mate, it ain't a problem .. I've already lost blood, but you're still going, so you're still streets ahead
and btw apologies are completely unnecessary, and I won't accept that one word description you gave yourself, as it's incorrect!
from downhere,
gt
they're my initials, fewer keystrokes!

gtho4,

Think I've got the answer.

Will try to check it out later this morning.

Bannockburn

gtho4,

I reckon you didn't give the poor old watchmaker enough time to finish his first cup of tea. You must have got to his shop almost as soon as he had opened the door.

You went in, he examined the clocks, he advised that they should be put under observation, whereupon he wound them up, set them to the correct time as he placed them on the shelf at 9:48 that Monday morning. He returned to the clocks at 14:36 the following Monday just as they were both striking 8 o'clock and he was able to calculate that clock "A" was running 45 minutes fast in a day and clock "B" was running 55 minutes slow in a day.

Although there are other bits and pieces to the problem that need to resolved, the heart of the problem lies in rationalising the equation

t( X + Y ) = 720

there "t" is the time in days that the clocks are left running, and "X" and "Y" are the minutes in each day that the clocks are running fast or slow.

It can be deduced that "t" must equal 7.2 with (X + Y ) equalling 100.

"X" and "Y" both being equal to 50 gives a time marginally before 9.30 am and is therefore unacceptable, but in working that out it determines for you that the clock running fast must take the 45 with the clock running slow taking the 55 as against the other way round.

That then gives the solution of 9:48 the previous Monday???

I'm keeping my fingers crossed??

Kindest Regards

Bannockburn

I have a message from Her Majesty, she's down here at the moment
mate, you've been knighted .. take a bow, and arise Sir Bannockburn!